# Question

Let $$\mathcal{B}$$ be the entire Borel set of $$\mathbb{R}$$. And let $$\chi_E$$ be the indicator function of the set $$E$$.

$$(X, m, \mu)$$; measure space, map $$f: X \to \mathbb{R}$$; measurable.

For $$A \in \mathcal{B}$$, we define $$\lambda(A) = \mu(f^{-1}(A)).$$

When $$g = \chi_E$$, $$E \in \mathcal{B}$$, the following equation transformation holds: $$\int_{\mathbb{R}} \chi_E\ d\lambda = \lambda(E) = \mu(f^{-1}(E)) \overset{?}{=} \int_{X} \chi_E \circ f\ d\mu.$$

Why is the third equal sign valid?

# What I know

$$\chi_E \circ f = \chi_E(f)$$,

$$\int_{X} \chi_E \circ f\ d\mu = \mu((\chi_E \circ f) \cap X)$$.

$$\chi_E \circ f$$ is not exactly $$\chi_E(f)$$, as the input set of $$\chi_E$$ is $$\mathbb{R}$$. Moreover, your formula $$\int_{X} \chi_E \circ f\ d\mu = \mu((\chi_E \circ f) \cap X)$$ doesn't make sense as $$(\chi_E \circ f) \cap X$$ is not a set as $$(\chi_E \circ f)$$ is a function and not a set.

More precisely, $$\chi_E \circ f$$ is a function from $$\mathbb{R}$$ to $$\mathbb{R}$$ defined by

$$\forall x \in \mathbb{R}, \quad (\chi_E \circ f) (x) = \chi_E(f(x)).$$

Therefore we can see that $$(\chi_E \circ f)= \chi_{f^{-1}(E)}$$. Indeed, for $$x \in \mathbb{R}$$, $$\begin{array}{lll} (\chi_E \circ f) (x) &=& \chi_E(f(x)) \\ &=& \left\{\begin{array}{lll}1 &\text{si }f(x) \in E \\ 0 &\text{otherwise} \end{array} \right. \\ &=& \left\{\begin{array}{lll}1 &\text{si }x \in f^{-1}(E) \\ 0 &\text{otherwise} \end{array} \right. \\ &=& \chi_{f^{-1}(E)}(x) \end{array}$$

Therefore we get $$\mu(f^{-1}(E)) = \displaystyle\int_X \chi_{f^{-1}(E)} d\mu = \displaystyle\int_X \chi_E \circ f d\mu.$$

• Thank you for letting us know about the mistake! I understood.
– ytnb
Dec 1, 2022 at 9:16